# Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

Clever student:

I know!

$x^{0}$ =  $x^{1-1}$ = $x^{1} x^{-1}$ = $\frac{x}{x}$ = $1$.

Now we just plug in x=0, and we see that zero to the zero is one!

Cleverer student:

No, you’re wrong! You’re not allowed to divide by zero, which you did in the last step. This is how to do it:

$0^{x}$$0^{1+x-1}$$0^{1} \times 0^{x-1}$$0 \times 0^{x-1}$$0$

which is true since anything times 0 is 0. That means that

$0^{0}$ = $0$.

Cleverest student :

That doesn’t work either, because if $x=0$ then

$0^{x-1}$ is $0^{-1} = \frac{1}{0}$

so your third step also involves dividing by zero which isn’t allowed! Instead, we can think about the function $x^{x}$ and see what happens as x>0 gets small. We have:

$\lim_{x \to 0^{+}} x^{x}$ = $\lim_{x \to 0^{+}} \exp(\log(x^{x}))$

= $\lim_{x \to 0^{+}} \exp(x \log(x))$

= $\exp( \lim_{x \to 0^{+} } x \log(x) )$

= $\exp( \lim_{x \to 0^{+} } \frac{\log(x)}{ x^{-1} } )$

= $\exp( \lim_{x \to 0^{+} } \frac{ \frac{d}{dx} \log(x) }{ \frac{d}{dx} x^{-1} } )$

= $\exp( \lim_{x \to 0^{+} } \frac{x^{-1}}{- x^{-2}} )$

= $\exp( \lim_{x \to 0^{+} } -x )$

= $\exp( 0)$

= $1$

So, since  $\lim_{x \to 0^{+}} x^{x}$ = 1, that means that $0^{0}$ = 1.

High School Teacher:

Showing that $x^{x}$ approaches 1 as the positive value x gets arbitrarily close to zero does not prove that $0^{0} = 1$. The variable x having a value close to zero is different than it having a value of exactly zero. It turns out that $0^{0}$ is undefined. $0^{0}$ does not have a value.

Calculus Teacher:

For all $x>0$, we have

$0^{x} = 0$.

Hence,

$\lim_{x \to 0^{+}} 0^{x} = 0$

That is, as x gets arbitrarily close to $0$ (but remains positive), $0^{x}$ stays at $0$.

On the other hand, for real numbers y such that $y \ne 0$, we have that

$y^{0} = 1$.

Hence,

$\lim_{y \to 0} y^{0} = 1$

That is, as y gets arbitrarily close to $0$, $y^{0}$ stays at $1$.

Therefore, we see that the function $f(x,y) = y^{x}$ has a discontinuity at the point $(x,y) = (0,0)$. In particular, when we approach (0,0) along the line with x=0 we get

$\lim_{y \to 0} f(0,y) = 1$

but when we approach (0,0) along the line segment with y=0 and x>0 we get

$\lim_{x \to 0^{+}} f(x,0) = 0$.

Therefore, the value of $\lim_{(x,y) \to (0,0)} y^{x}$ is going to depend on the direction that we take the limit. This means that there is no way to define $0^{0}$ that will make the function $y^{x}$ continuous at the point $(x,y) = (0,0)$.

Mathematician: Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true.

Let’s consider the problem of defining the function $f(x,y) = y^x$ for positive integers y and x. There are a number of definitions that all give identical results. For example, one idea is to use for our definition:

$y^x$ := $1 \times y \times y \cdots \times y$

where the y is repeated x times. In that case, when x is one, the y is repeated just one time, so we get

$y^{x}$ = $1 \times y$.

However, this definition extends quite naturally from the positive integers to the non-negative integers, so that when x is zero, y is repeated zero times, giving

$y^{0}$ = $1$

which holds for any y. Hence, when y is zero, we have

$0^0 = 1$.

Look, we’ve just proved that $0^0 = 1$! But this is only for one possible definition of $y^x$. What if we used another definition? For example, suppose that we decide to define $y^x$ as

$y^x$ := $\lim_{z \to x^{+}} y^{z}$.

In words, that means that the value of $y^x$ is whatever $y^z$ approaches as the real number z gets smaller and smaller approaching the value x arbitrarily closely.

[Clarification: a reader asked how it is possible that we can use $y^z$ in our definition of $y^x$, which seems to be recursive. The reason it is okay is because we are working here only with $z>0$, and everyone agrees about what $y^z$ equals in this case. Essentially, we are using the known cases to construct a function that has a value for the more difficult x=0 and y=0 case.]

Interestingly, using this definition, we would have

$0^0$ = $\lim_{x \to 0^{+}} 0^{x}$ = $\lim_{x \to 0^{+}} 0$ = $0$

Hence, we would find that $0^0 = 0$ rather than $0^0 = 1$. Granted, this definition we’ve just used feels rather unnatural, but it does agree with the common sense notion of what $y^x$ means for all positive real numbers x and y, and it does preserve continuity of the function as we approach x=0 and y=0 along a certain line.

So which of these two definitions (if either of them) is right? What is $0^0$ really? Well, for x>0 and y>0 we know what we mean by $y^x$. But when x=0 and y=0, the formula doesn’t have an obvious meaning. The value of $y^x$ is going to depend on our preferred choice of definition for what we mean by that statement, and our intuition about what $y^x$ means for positive values is not enough to conclude what it means for zero values.

But if this is the case, then how can mathematicians claim that $0^0=1$? Well, merely because it is useful to do so. Some very important formulas become less elegant to write down if we instead use $0^0=0$ or if we say that $0^0$ is undefined. For example, consider the binomial theorem, which says that:

$(a+b)^x$ = $\sum_{k=0}^{\infty} \binom{x}{k} a^k b^{x-k}$

where $\binom{x}{k}$ means the binomial coefficients.

Now, setting a=0 on both sides and assuming $b \ne 0$ we get

$b^x$

= $(0+b)^x$ = $\sum_{k=0}^{\infty} \binom{x}{k} 0^k b^{x-k}$

= $\binom{x}{0} 0^0 b^{x} + \binom{x}{1} 0^1 b^{x-1} + \binom{x}{2} 0^2 b^{x-2} + \hdots$

= $\binom{x}{0} 0^0 b^{x}$

= $0^0 b^{x}$

where, I’ve used that $0^k = 0$ for k>0, and that  $\binom{x}{0} = 1$. Now, it so happens that the right hand side has the magical factor $0^0$. Hence, if we do not use $0^0 = 1$ then the binomial theorem (as written) does not hold when a=0 because then $b^x$ does not equal $0^0 b^{x}$.

If mathematicians were to use $0^0 = 0$, or to say that $0^0$ is undefined, then the binomial theorem would continue to hold (in some form), though not as written above. In that case though the theorem would be more complicated because it would have to handle the special case of the term corresponding to k=0. We gain elegance and simplicity by using $0^0 = 1$.

There are some further reasons why using $0^0 = 1$ is preferable, but they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more “natural” to mathematicians. The choice is not “right”, it is merely nice.

This entry was posted in -- By the Mathematician, Math, Philosophical. Bookmark the permalink.

### 1,158 Responses to Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree?

1. STEPHAN H. says:

So rather than carve out exceptions to some of our favorite theorems and definitions, let us consider the following formulation in algebra textbooks (at least going forward):

x^0 ≡ 1 (for any real number x)
0^x ≡ 0 (for any positive number x)

I leave it as a challenge to any remaining quibblers to show how the first definition above leads to any contradiction by rigorous logical reasoning.

2. Alan Feldman says:

1) The indeterminate form applies to categories of limits, not values.

Consider the limit

lim(h->oo) (1+x/h)^h

This is a limit of the form 1^oo. The answer is e^x. This does not mean that 1^oo is e^x. Similarly, lack of a unique limit for 0^0 does not imply that 0^0 is undefined.

(Hmmm, I thought I’ve read a number of times that limits are always unique!)

The “indeterminate form” 1^oo is simply being used as a sensible symbol or label for the category of limits where the base approaches 1 and the exponent approaches oo as the limit is taken.

2) Even if you still insist on using limits, you can only do that if the function is continuous. Why does 0^x need to be continuous? It’s already a problem for x<0, and probably for imaginary numbers, too.

3) These are definitions. As I've asked before, how do you derive y^x = e^(x ln y)?

Well, you'd like things to obey the exponent laws. So you show that the log definition does so. But consider y^m y^n = y^(m+n) for even roots.

For 4^(1/2) you can say it's 2 or -2. Indeterminate!!! You sensibly omit the negative root and get on with things. Physics is quite happy with it too.

(Furthermore, my analysis book (_Introduction to Analysis_ by Rosenlicht) says it is defined, not derived.)

For 0^0: To satisfy the exponent rules, 0^0 has to be 0 or 1. We sensibly throw out 0, because 1 is immensely useful and 0 isn't. Power series, mappings, derivative of x^n, and more. It's 1. 0^0 = 1 works with all of these, and 0 doesn't. It's 1.

Why is one any less intrinsic or derivable than the other? They're really just definitions.

4) As Howard pointed out, what is the sense of saying, "Whenever it comes up, substitute 1, but it really is undefined"?

5) Please read the posts by Howard Ludwig, betaneptune, and me and let us know where we went wrong.

3. Alan Feldman says:

Regarding calls to close this thread:

I welcome arguments showing where Howard Ludwig, betaneptune, I, and now STEPHAN H. have gone wrong.

I also welcome new arguments. Please, please, please — no more division by zero arguments. I’ve answered that too many times. Thank you.

Lacking either, I’m pretty much done. (I reserve the right to reverse this!)

4. Richard says:

@Alan Feldman

When I suggested that it’s getting to be time to wrap up the discussion, I didn’t mean officially closing it to any further postings.  I just meant that we all seem to be making the same points and stating the same arguments over and over again, and I don’t see much point in doing that.  But there’s always the possibility someone will bring up a previously unconsidered aspect of this, or a new contributor may have something to say.

=========================

@STEPHAN H.  has written:

“Let us agree, for the sake of convenience and epistemological honesty, that it is axiomatic (a useful definition and not a theorem) that  0º ≡ 1”

*If I am interpreting you correctly*, then we may have something we can actually agree on.  I agree that, for most of the situations where 0^0 comes up, it is convenient to consider it to be 1.   Calling it “a useful definition and not a theorem” to me implies that we’re defining it to be 1 as a matter of practicality rather than claiming that it can be rigorously proven to be 1 just as surely as  a^(1/2)  can be rigorously proven to be the square root of (a).  If that’s more or less what you mean, then we’re basically in agreement, at least concerning that statement.

5. The Physicist says:

This is getting a bit circular for those involved, and difficult to follow for everyone else.
For the sake of casual readers, I’m going to close this to comments that rehash the same ideas. If anyone would like to make their point again, I’ll consider short, clear comments with explicit, directly supporting, linked references.

6. Alan Feldman says:

The definition of x^n is quite clear when n is a positive integer. Saying x^n = exp(n log x) for x > 0 and 0^n = 0 for n > 0 is a matter of “extending” the definition of x^n. So these are not proofs, derivations, or theorems. Reference:

108.243.42.146:8080/get/pdf/1039

pp. 130-131

That puts them on an equal standing with 0^0 = 1, except for the controversy about it.

7. Dennis Sloane says: